consider the chemical equations shown here.\n...
consider the chemical equations shown here.\np4(s) + 3o2(g) → p4o6(s) δh1 = -1,640.1 kj\np4o10(s) → p4(s) + 5o2(g) δh2 = 2,940.1 kj\nwhat is the overall enthalpy of reaction for the equation shown below?\nround the answer to the nearest whole number.\np4o6(s) + 2o2(g) → p4o10(s)
Answer
# Explanation:
## Step1: Revertir la primera ecuación
$$
\begin{align*}
P_4O_6(s)&\to P_4(s)+3O_2(g)\\
\Delta H_1'& = 1640.1\ kJ
\end{align*}
$$
## Step2: Revertir la segunda ecuación
$$
\begin{align*}
P_4(s)+5O_2(g)&\to P_4O_{10}(s)\\
\Delta H_2'&=- 2940.1\ kJ
\end{align*}
$$
## Step3: Sumar las ecuaciones revertidas
$$
\begin{align*}
P_4O_6(s)+P_4(s)+5O_2(g)&\to P_4(s)+3O_2(g)+P_4O_{10}(s)\\
P_4O_6(s)+2O_2(g)&\to P_4O_{10}(s)
\end{align*}
$$
$$
\begin{align*}
\Delta H&=\Delta H_1'+\Delta H_2'\\
&=1640.1+( - 2940.1)\\
&=-1300\ kJ
\end{align*}
$$
# Answer:
$-1300$